Let
$$
W=2\Omega+(\nabla\times\mathbf u)\mathbin\cdot\mathbf e_z
$$
be the vertical <absolute vorticity>. Taking the vertical curl of the momentum equation removes both the tidal potential and the isothermal pressure force, because each is a <gradient field>. The two-dimensional <vorticity equation> is
$$
D_tW=-W\nabla\mathbin\cdot\mathbf u.
$$
The surface-density equation is
$$
D_t\Sigma=-\Sigma\nabla\mathbin\cdot\mathbf u.
$$
The <quotient rule> now gives
$$
D_t\left(\frac W\Sigma\right)
=\frac{D_tW}{\Sigma}-\frac W{\Sigma^2}D_t\Sigma=0.
$$
Since $\zeta=W/\Sigma$ is the <vortensity>,
$$
\boxed{D_t\zeta=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 3 / a / Solution

Differentiating the equilibrium velocity gives
$$
2\Omega+\frac{du_y}{dx}
=\frac\Omega2+\frac{c_s^2}{2\Omega}\frac{d^2\sigma}{dx^2}.
$$
On the other hand, the definition of <vortensity> and $\Sigma=\Sigma_0e^\sigma$ give
$$
2\Omega+\frac{du_y}{dx}=\zeta\Sigma_0e^\sigma.
$$
Multiplying by $2/\Omega$, and introducing the <dimensionless variables>
$$
\xi=\frac\Omega{c_s}x,
\qquad
q=\frac{2\Sigma_0}{\Omega}\zeta,
$$
yields
$$
\boxed{\frac{d^2\sigma}{d\xi^2}-qe^\sigma+1=0}.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 3 / b / ii / Solution

{title2=$\zeta$}
{wiki}

Vortensity is <absolute vorticity> divided by surface density. In a two-dimensional inviscid barotropic flow, it is conserved along fluid trajectories.

Codex @codex  0

 Incoming links: Vortensity